Consider the two sets : A={m in R: | vedclass

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Question

$A: D \geq 0$
$\Rightarrow \quad(m+1)^{2}-4(m+4) \geq 0$
$\Rightarrow \quad m^{2}+2 m+1-4 m-16 \geq 0$
$\Rightarrow \quad m^{2}-2 m-15 \geq 0$
$\R

Vedclass · Accepted Answer

$A-B=(-\infty,-3) \cup(5, \infty)$

Vedclass · Answer

$A: D \geq 0$
$\Rightarrow \quad(m+1)^{2}-4(m+4) \geq 0$
$\Rightarrow \quad m^{2}+2 m+1-4 m-16 \geq 0$
$\Rightarrow \quad m^{2}-2 m-15 \geq 0$
$\Rightarrow \quad(m-5)(m+3) \geq 0$
$\Rightarrow \quad m \in(-\infty,-3] \cup[5, \infty)$
$	herefore \quad A=(-\infty,-3] \cup[5, \infty)$
$B=[-3,5)$
$A-B=(-\infty,-3) \cup[5, \infty)$
$A \cap B=\{-3\}$
$B-A=(-3,5)$
$A \cup B=R$

Consider the two sets : $A=\{m \in R:$ both the roots of $x^{2}-(m+1) x+m+4=0$ are real $\}$ and $B=[-3,5)$ Which of the following is not true?

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